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Triss [41]
3 years ago
13

How would you solve this equation? Please show steps

Mathematics
1 answer:
Serhud [2]3 years ago
4 0
Simple....

you have: 3x= \sqrt{12-12x}

So...first you know you're trying to find x..

But, you have a square root on one side...to remove this you'd have to square both sides--->>>

(3x)^{2} = \sqrt{12-12x}  ^{2}

Leaving you with....

9x^{2} =12-12x

Move the terms over and set it equal to zero.

-->>

9x^{2}+12x-12=0

Factor out a 3...

3( 3x^{2} +4x-4=0)

What multiplies to -12 and adds to 4?

6*-2=-12

6+-2=4

Leaving you with...

3(3x-2)(x+2)=0

Remember you're solving for 0....

3x-2=0

3x-2=0
    +2  +2

3x=2

\frac{3x}{3} = \frac{2}{3}

x=
\frac{2}{3}

Then-->>

x+2=0

x+2=0
  -2   -2

x=-2

Thus, your answer.
 


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I don't think this is right

4 0
3 years ago
Two vectors A⃗ and B⃗ are at right angles to each other. The magnitude of A⃗ is 5.00. What should be the length of B⃗ so that th
creativ13 [48]

Answer:

Length of B is 7.4833

Step-by-step explanation:

The vector sum of A and B vectors in 2D is

C=A+B=(a_1+b_1,a_2+b_2)

And its magnitude is:

C=\sqrt{(a_1+b_1)^2+(a_2+b_2)^2} =9

Where

a_1=Asinx

a_2=Acosx

b_1=Bsin(x+90)

b_2=Bcos(x+90)

Using the properties of the sum of two angles in the sin and cosine:

b_1=Bsin(x+90)=B(sinx*cos90+sin90*cosx)=Bcosx

b_2=Bcos(x+90)=B(cosx*cos90-sinx*sin90)=-Bsinx

Sustituying in the magnitud of the sum

C=\sqrt{(Asinx+Bcosx)^2+(Acosx-Bsinx)^2} =9

C=\sqrt{A^2sin^2x+2ABsinxcosx+B^2cos^2x+A^2cos^2x-2ABsinxcosx+B^2sin^2x} =9

C=\sqrt{A^2(sin^2x+cos^2x)+B^2(cos^2x+sin^2x)}

C=\sqrt{A^2+B^2} =9

Solving for B

A^2+B^2 =9^2

B^2 =9^2-A^2

Sustituying the value of the magnitud of A

B^2=81-5^2=81-25=56

B= 7. 4833

8 0
3 years ago
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